Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define ll long long
#define fop(i,m,n) for(ll i=m;i<n;i++)
using namespace std;
int main(){
	ll n;
	cin >> n;
	vector<double> dp(1,1);
	fop(i,0,n){
		double x;
		cin >> x;
		vector<double> dpp(i+2);
		fop(j,0,i+1){
			dpp[j] += dp[j] * (1 - x);
			dpp[j+1] += dp[j] * x;
		}
		swap(dp,dpp);
	}
	double ans = 0;
	fop(i,0,n+1){
		if(i > n-i) ans += dp[i]; 
	}
	cout << fixed << setprecision(12) << ans;
}